Does Charge conjugation change the spin momentum? Charge conjugation of a particle of spin momentum up is an anti-particle of spin momentum up or down?
 A: Spin is even under parity, even under charge reversal, and odd under time reversal.
This symmetry is an important argument for why spin-half particles without internal structure may have magnetic dipole moments (such a moment being changing sign under all three of $P,C,T$) but not permanent electric dipole moments (which would change sign under $P$ and $C$ but not $T$).  We observe that $T$ is approximately a good symmetry in nature.  The term $\vec\sigma\cdot\vec B$, by multiplication, does not change sign under $T$ or $CP$, and so may appear in a $CP$-even Hamiltonian; terms like $\vec\sigma\cdot\vec E$ are odd under $CP$ and $T$ and are forbidden except a the small level that $CP$ symmetry is broken.
A: As a consequence of Wigner's definition of elementary particle (relying upon the theory of irreducible unitary representations of Poincaré group), the Hilbert space of the states of a particle has always this structure
$$H= H_{orbital}\otimes H_{spin} \otimes H_{internal}\:.$$ 
The orbital part is that concerning position, momentum and so on. It is always of the form $L^2(\mathbb R^3)$ referred to the measure either  $d^3x$, or $d^3p$ or some covariant version of the latter (all these representation are unitarily equivalent). $H_{spin}$ is the spin/ helicity space, it is isomorphic to $\mathbb C^j$ for a suitable integer $j$ (if the particle is massive $j=2s+1$, where $s$ is the spin of the particle).
Poincaré group acts in $H_{orbital}\otimes H_{spin}$.
The space $H_{internal}$ describes all internal degrees of freedom of the particle, in particular the electrical charge. The charge conjugation operator $C$ is a unitary involutive operator acting in $H_{internal}$. More precisely it has  the form $I_{orbital}\otimes I_{spin}\otimes C$. 
It should be clear that the charge conjugation operation cannot change the spin state of the particle.
